The equation of a circle $C$ is $x^2+y^2-6x+6y+17 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2-6x) + (y^2+6y) = -17$ $(x^2-6x+9) + (y^2+6y+9) = -17 + 9 + 9$ $(x-3)^{2} + (y+3)^{2} = 1 = 1^2$ Thus, $(h, k) = (3, -3)$ and $r = 1$.